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| classes:2009:fall:phys4101.001:q_a_0930 [2009/09/29 22:03] – yk | classes:2009:fall:phys4101.001:q_a_0930 [2009/10/05 20:06] (current) – yk |
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| === Anaximenes - 19:20 - 09/28/09 === | === Anaximenes - 19:20 - 09/28/09 === |
| As the professor said during class, <math>E=\(n + \frac{1}{2}\)\hbar\omega</math> starting from 0 makes it more clear what the ground state energy is than <math>E=\(n - \frac{1}{2}\)\hbar\omega</math> starting at n = 1. | As the professor said during class, <math>E=\(n + \frac{1}{2}\)\hbar\omega</math> starting from 0 makes it more clear what the ground state energy is than <math>E=\(n - \frac{1}{2}\)\hbar\omega</math> starting at n = 1. |
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| | ==== Blackbox - 10:10 - 10/04/09 ==== |
| | If it starts with n=0 for the harmonic oscillator, what about the ground state for infinite square well? |
| | As you know this <math>E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}</math>, Why doesn't it begin with n=0? |
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| ====John Galt 18:19 9/28 (can you include date/time next time, John?)==== | ====John Galt 18:19 9/28 (can you include date/time next time, John?)==== |
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| I don't think anyone is suggesting that humanity give up on finding some reason for characteristics of particles to be indeterminate (maybe look into the source of the Schrodinger equation?), but (a) to make progress, we have to accept propositions at least as suppositions, and (b) no matter how much we ask "why are things like this?", eventually, the answer will have to be "they just are." Eventually, there won't be another answer. This isn't to suggest that we should stop searching; I mean that a statement that it looks like things "just are" has never stopped scientists from continuing to ask "why;" things have always seemed like they "just were." The fact that it seems like particles' location and momentum are indeterminate doesn't mean we won't keep exploring; people all over the world conduct research in the field of quantum mechanics. | I don't think anyone is suggesting that humanity give up on finding some reason for characteristics of particles to be indeterminate (maybe look into the source of the Schrodinger equation?), but (a) to make progress, we have to accept propositions at least as suppositions, and (b) no matter how much we ask "why are things like this?", eventually, the answer will have to be "they just are." Eventually, there won't be another answer. This isn't to suggest that we should stop searching; I mean that a statement that it looks like things "just are" has never stopped scientists from continuing to ask "why;" things have always seemed like they "just were." The fact that it seems like particles' location and momentum are indeterminate doesn't mean we won't keep exploring; people all over the world conduct research in the field of quantum mechanics. |
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| ====Andomeda==== | ====Andomeda 09/28 23:51==== |
| when it says "the quantum mechanical wave function travels at half the speed of the particle it is supposed to represent!" on page 60, I understand why the velocities are different but which one (classical velocity or quantum velocity) is the speed that can not be more than the speed of light? | when it says "the quantum mechanical wave function travels at half the speed of the particle it is supposed to represent!" on page 60, I understand why the velocities are different but which one (classical velocity or quantum velocity) is the speed that can not be more than the speed of light? |
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| The classical velocity is the group velocity of the wave packet and the quantum velocity is the phase velocity, and from what I understand the phase velocity cannot be faster than the speed of light. However, I don't think what Griffiths was talking about applies to relativistic particles. | The classical velocity is the group velocity of the wave packet and the quantum velocity is the phase velocity, and from what I understand the phase velocity cannot be faster than the speed of light. However, I don't think what Griffiths was talking about applies to relativistic particles. |
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| | === Can 11:01pm 9/28 === |
| | For Chavez, phase velocity could be larger than speed of light. And you are right about the first half, group velocity usually denotes the velocity of the wavepacket or particle velocity, which is also the velocity of energy propagation. However phase velocity is the change of phase modulation inside the wavepacket. |
| ==== joh04684 - 7:17 - 9/29/09 ==== | ==== joh04684 - 7:17 - 9/29/09 ==== |
| For the free particle, where exactly does the motivation behind adding in <math>\phi(k)</math> into the equation of the wave? Is it just some mathematical trick to make everything work out alright? It looks like they might be replacing A with that, so they no longer have just a constant for normalization, but rather a function of the variable we're interested in anyhow, is that correct? | For the free particle, where exactly does the motivation behind adding in <math>\phi(k)</math> into the equation of the wave? Is it just some mathematical trick to make everything work out alright? It looks like they might be replacing A with that, so they no longer have just a constant for normalization, but rather a function of the variable we're interested in anyhow, is that correct? |
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| | === Daniel Faraday 7am 9-30 === |
| | I believe the <math>\phi(k)</math> in the free particle is analogous to Cn for the square well. In the square well, the stationary states are a //discrete// set, and each stationary state has an index n. So, the linear combination of the stationary states is a //summation//, and each element in the sum has its own coefficient, Cn. But with a free particle, the stationary states are a //continuous// set, since k can have any value. So the general solution is an //integral// of all of these stationary states. |
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| | This means that instead of a Cn, which is like a function that is only defined for each n, we have a continuous function of coefficients, <math>\phi(k)</math>. I think of it as though there’s one value of <math>\phi(k)</math> for each one of the possibly infinite number of waves that are added together to make up the wave packet. |
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| | I think this is another way of saying that in a square well any wavefunction is a linear combination of discrete stationary states, whereas for a free particle, any localized wave packet is a sum (possibly infinite) of lots of waves of different frequencies, which interfere with each other everywhere in space to completely cancel out except where the wave packet is. |
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| | At least that’s what I understood from my TA when I asked him this exact question yesterday. |
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| ==== joh04684 - 8:14 - 9/29/09 ==== | ==== joh04684 - 8:14 - 9/29/09 ==== |
| Working through last-year's exam, I couldn't find any examples of when we'd done an infinite square well going from -a/2 to a/2, instead most of our examples (and in the book) go from 0 to L, so I was trying to derive the general <math>\psi_n(x)</math> solutions. Is it correct to use the form <math> \psi_n (x) = \sqrt{\frac{2}{a}}\sin{\frac{2n\pi}{a}x)</math>, and <math>E_n = \frac{2n^2\pi^2\hbar^2}{ma^2}</math>, where the only differences between a well from 0 to L and -a/2 to a/2 is inside the sin function (2nPi/a instead of nPi/a) and in energy (where you end up with a 2^2/a^2 instead of 1/L^2), but the normalization constant remains the same? Or is there some other simple intuitive way of formulating a solution to this? | Working through last-year's exam, I couldn't find any examples of when we'd done an infinite square well going from -a/2 to a/2, instead most of our examples (and in the book) go from 0 to L, so I was trying to derive the general <math>\psi_n(x)</math> solutions. Is it correct to use the form <math> \psi_n (x) = \sqrt{\frac{2}{a}}\sin{\frac{2n\pi}{a}x)</math>, and <math>E_n = \frac{2n^2\pi^2\hbar^2}{ma^2}</math>, where the only differences between a well from 0 to L and -a/2 to a/2 is inside the sin function (2nPi/a instead of nPi/a) and in energy (where you end up with a 2^2/a^2 instead of 1/L^2), but the normalization constant remains the same? Or is there some other simple intuitive way of formulating a solution to this? |
| ===Andromeda=== | ===Andromeda 09/29 20:42=== |
| i think the energy equation remains the same but the time independent schrodinger equation will either be <math>A_n\sin k_nx</math> for //n//=even integers or <math>B_n\cos k_nx</math> for //n//=odd integers due to the slightly different boundary condition. | i think the energy equation remains the same but the time independent schrodinger equation will either be <math>A_n\sin k_nx</math> for //n//=even integers or <math>B_n\cos k_nx</math> for //n//=odd integers due to the slightly different boundary condition. |
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| ==== --------------------------------- ==== | ==== joh04684 - 10:15 - 9/30/09 ==== |
| | Up until now all of our dealings in quantum mechanics have been treating mass as constant. My question is, how do you work quantum mechanics when mass also varies in time or space, such as a decaying particle? |
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| | === Esquire 10:16 9/30/09 === |
| | I have also pondered this question. In short, quantum field theory is required to analyze quantum systems of variable mass. |
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| | === Pluto 4ever 6:27pm 9/30/09 === |
| | I also agree. You only need a single mass to analyze how quantum systems (e.g. atoms, light) work which applies to other masses of these systems. So whether it is the Schrodinger Equation or the Heisenberg Uncertainty Principle, you are more concerned about the behavior of the particle. |
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| | ==== Malmx026 9/30 7:45pm ==== |
| | My question is about the Pauli Exclusion Principle (a little more general). At what length do two fermions become close enough to say that they cannot have the same quantum state? More specifically this is shown in materials when the distance between fermions get close and the discrete band energies become spread out. Do the particles interact? |
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| | ==== Malmx026 9/30 8:45pm ==== |
| | In problem 2.1 it is asked to show that imaginary energy is not possible because the wave function is not normalizable for all time, but if time is allowed to be imaginary then this isn't the case. Is imaginary time necessary to describe particles in any physical situation? Thanks |
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